The core trick is prefix sums mod k, store the first time you see each remainder and check if you've seen it before with enough distance between indices.
Use prefix sums and modular arithmetic: compute prefix sums modulo k and check if any remainder repeats within a distance of at least 2. If a remainder repeats, the subarray between those indices has a sum divisible by k; ensure its length is at least 2 by tracking the earliest index for each remainder.
Pro tip: Handle edge cases explicitly: k=0 (though typically k>0), negative numbers (modulo operation in some languages returns negative), and the length constraint. Also, mention that the solution runs in O(n) time and O(k) space, which is optimal.
Confirm that k is positive, array can contain negative numbers, and subarray must be contiguous with length >= 2. Ask about expected input size to determine if O(n) is necessary.
State that if two prefix sums have the same remainder modulo k, the sum of the elements between them is a multiple of k. This is because (prefix[j] - prefix[i]) % k == 0.
Iterate through the array, compute running sum modulo k, and store the first index where each remainder occurs. If a remainder repeats and the distance between indices is at least 2, return true.
Ensure modulo results are non-negative (e.g., in Python use % k, in Java use Math.floorMod). Also, initialize the map with remainder 0 at index -1 to handle subarrays starting from index 0.
State time complexity O(n) and space O(min(n, k)). Walk through a small example to verify, including cases with negative numbers and no valid subarray.
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